Question

A point mass \(m\) is a distance \(d\) from the center of a thin spherical shell of mass \(M\) and radius \(R\). The magnitude of the gravitational force on the point mass is given by the integral \(F(d)=\frac{G M m}{4 \pi} \int_{0}^{2 \pi} \int_{0}^{\pi} \frac{(d-R \cos \varphi) \sin \varphi}{\left(R^{2}+d^{2}-2 R d \cos \varphi\right)^{3 / 2}} d \varphi d \theta\), where \(G\) is the gravitational constant. a. Use the change of variable \(x=\cos \varphi\) to evaluate the integral and show that if \(d>R,\) then \(F(d)=\frac{G M m}{d^{2}},\) which means the force is the same as if the mass of the shell were concentrated at its center. b. Show that if \(d

Short answer

Answer: When the point mass \(m\) is outside the shell (i.e., \(d>R\)), the gravitational force acting on it is given by \(F(d) = \frac{GMm}{d^2}\), where \(G\) is the gravitational constant. When the point mass is inside the shell (i.e., \(d<R\)), there is no gravitational force acting on it, meaning \(F = 0\).

Step-by-step solution

Change of variable

Let's change the variable from \(\varphi\) to \(x=\cos \varphi\). We will need the corresponding Jacobian to do this transformation. The Jacobian of this transformation is given by: $$\frac{dx}{d\varphi} = \frac{d(\cos \varphi)}{d\varphi} = -\sin \varphi$$ Thus, $$d\varphi = -\frac{dx}{\sin \varphi}$$

Update the integral

Now let's substitute the variable change and its Jacobian into the integral: $$F(d) = \frac{GMm}{4\pi} \int_{0}^{2\pi} \int_{-1}^{1} \frac{(d-Rx) \sin \varphi}{\left(R^{2}+d^{2}-2Rd x\right)^{3/2}}(-dx)d\theta$$

Simplify the integral

The integral can be simplified to: $$F(d) = \frac{GMm}{2\pi} \int_{0}^{2\pi} \int_{-1}^{1} \frac{(d-Rx)}{\left(R^{2}+d^{2}-2Rd x\right)^{3/2}} dx d\theta$$

Solve the inner integral

We now integrate with respect to \(x\): $$\int_{-1}^{1} \frac{(d-Rx)}{\left(R^{2}+d^{2}-2Rd x\right)^{3/2}} dx = \int_{-1}^1 \frac{1}{(R^{2}+d^{2}-2Rd x)^{1/2}} - \frac{R}{R^{2}+d^{2}-2Rd x}$$ The left integral can be evaluated using the substitution \(u = R^2+d^2-2Rdx\), and the right integral is an odd function, so its integral is zero.

Evaluate the outer integral

Now we just need to evaluate the outer integral: $$F(d) = \frac{GMm}{2\pi} \cdot 2\pi \cdot \frac{1}{d^2} = \frac{GMm}{d^2}$$ #b. Evaluating the integral when \(d<R\)#

Show that the integral is zero

If \(d<R\), notice that the integrand is an odd function with respect to \(x\): $$f(x) = \frac{(d-Rx)}{\left(R^{2}+d^{2}-2Rd x\right)^{3/2}} = -f(-x)$$ Since the limits of integration are symmetric, the integral evaluates to zero.

Final answer

Therefore, when the point mass is inside the shell, \(F = 0\), which means there is no gravitational force.

Key concepts

Spherical Shell Theorem

The Spherical Shell Theorem is a fascinating aspect of gravitational physics. It states that a spherically symmetric shell of mass exerts no net gravitational force on a point mass located inside it. What's compelling is that the gravitational force outside the shell acts as though all the shell's mass is concentrated at a single point at its center.
This theorem simplifies the calculation of gravitational forces in systems where spheroid shells are involved, like planets and stars.

• When a point mass is outside the shell (\(d > R\)), the force is as if the entire mass is at the shell's center.
• Inside the shell (\(d < R\)), the net gravitational force is zero, thanks to the symmetry and uniformity of the shell.

Understanding this theorem is crucial in astrophysics and helps explain why gravity on Earth's surface is nearly uniform.

Integration Techniques

Integration is a powerful mathematical tool for calculating complex quantities like gravitational forces that involve continuous distributions of mass. In our exercise, integration allows us to sum up infinitesimal forces around a spherical shell.
We initially encounter a double integral because both the elevation angle \(\varphi\) and azimuthal angle \(\theta\) are involved. However, using spherical symmetry, the \(\theta\) integration simplifies, and the focus is on the radial part of the integral.

• Using symmetry, we reduce multiple dimensions to a manageable form.
• We break down the integral into more straightforward parts, often tackling the radial or angular components separately.

Integration is more tractable when these techniques reduce complexity while retaining the integral's essence. Mastering these techniques deeply enhances problem-solving skills in calculus and physics.

Variable Substitution

Variable substitution is an essential technique that simplifies evaluating integrals by transforming one variable into another more suitable for integration. In our exercise, we substitute \(x = \cos \varphi\) to manage the integral efficiently.
This substitution changes the variables from spherical to Cartesian-like ones, making the problem more accessible. With it, the integral bounds adjust from angles to simple intervals (\(-1\) to \(1\)) along \(x\).

• Jacobian transformation is essential here. It helps account for differential changes during substitution.
• This technique brings complex angular measure into a linear form, streamlining the solution.

Variable substitution can greatly simplify integrals, making it easier to find solutions in complex scenarios such as gravitational fields involving angles.